M12-11

I see there is some uncertainty over the problem wording; the intention of the problem is asking:

\(1.2^n\geq 2\), (solving for n, where n is the lowest possible integer)

I haven't seen this wording on official questions, but on this question, "before" means "how many years will it take for the population to at least double"? If "before" is taken literally as "prior to doubling", then 0 or 1 or 2 would also be acceptable responses, which doesn't make sense.


2 solutions below:

1) A reliable way to approach percent change over a short period of years (essentially, compound interest) is a table, picking the number 100 as a starting point:

Year 0: 100
After 1 year: 120
After 2 years: 120*1.2 = 144
After 3 years: 144*1.2 = 172.8 (a quick way to add 20% is add 2*10% --> add 2*14.4 to 144 --> 172.8
After 4 years: 172.8*1.2 = 207.36
Here, we can save time by rounding and approximating -- we just need to know Yes/No whether it's greater than 200 (just like Data Sufficiency, it's best to avoid unnecessary calculations)
Round to 173 and add 2*10% --> 173 + 2*17.3 > 200? --> Clearly yes (173 + 34.6 = 207.6)

Answer is B. 4


2) Using fractions instead of decimals:

In many cases, it's a helpful habit to use fractions:
Translate "increase (grows) by 20%" to "multiply by \(\frac{6}{5}\)"

Glance at the answer choices and try 3 years first:

If you have your cubes memorized, \((\frac{6}{5})^3=\frac{216}{125}<2.\) Less than 2, so not the answer. \((2 = \frac{250}{125}\))


Next, try 4 years: \(\frac{216}{125}*\frac{6}{5}=\frac{1296}{625} > 2.\) Greater than 2, so our answer is B. 4 (\(2 = \frac{1250}{625}\))